Chapter 3: The Laplace Transform3.1. Definition and Basic Properties

。 Objective of Laplace transform

-- Convert differential into algebraic equations

○ Definition 3.1: Laplace transform

s.t. converges

s, t : independent variables

1

,s0

[ ]( ) ( )stL f s e f t dt

[ ], [ ], [ ]F L f G L g H L h ＊ Representation:

。Example 3.2:

2

( ) sing t t

0 0

02 2

( ) ( ) sin

cos sin 1 | ,

1 1

st st

st st

G s e g t dt e t dt

e t se ts

s s

0

sin .ste tdt

Let , sin

, cos

st

st

u e dv tdt

du se dt v t

0 0

sin cos cosst st ste tdt e t se tdt

From udv uv vdu

Consider

3

Let , cos .

, sin

st

st

u e dv tdt

du se dt v t

0 0

sin cos cosst st ste tdt e t se tdt

2

0

2 20

(1 ) sin cos sin

cos sin 1sin

01 1

st st st

st stst

s e tdt e t s te

e t s tee tdt

s s

0

cos (sin sin )st st ste t s te se tdt

2

0

cos sin sinst st ste t s te s e tdt

* Not every function has a Laplace transform.

In general, can not converge

。 Example 3.1:

4

0( ) ste f t dt

s

( ) , : any real numberatf t e a

( )

0 0 0

( )0

( ) ( )

1 1 | ,

st st at a s t

a s t

F s e f t dt e e dt e dt

e s aa s s a

○ Definition 3.2.: Piecewise continuity (PC)

f is PC on if there are finite points

s.t.

and are finite

5

[ , ]a b

1 2 na t t t b

1 1 2 is continuous on ( , ) ( , ) ( , )nf a t t t t b

lim ( ), lim ( )

lim ( ), lim ( )j

j

t a t t

t t t b

f t f t

f t f t

i.e., f is continuous on [a, b] except at finite points, at each of which f has finite one-sided limits

6

2 0 2t

22 e.g., ( )

2 31

3 41

t

tf x

t

t

If f is PC on [0, k], then so is and

exists0

( )k ste f t dt

( )ste f t

◎ Theorem 3.2: Existence of

f is PC on

If

Proof:

7

[ ]L f

[0, ] 0k k , , s.t. ( ) , > 0btM b f t Me t

0then ( ) converges ste f t dt s b

[ ]( ) is defined for > L f s s b

( ) ( ) , ( )bt st b s tf t Me e f t Me

( ) ( )

0 0 0

0

( )

( ) converges

st b s t s b t

st

M Me f t dt Me dt e s b

s b s b

e f t dt

0 0 0( ) ( ) ( ) convergesst st ste f t dt e f t dt e f t dt

8

1

2

1

2

0

e.g., 0, ( ) is not PC on any

[0, ] since limt

t f t t

k t

2

2

1

2

0

1/2

0

0

[ ]( )

2 (where )

2 (where )

/

st

sx

z

L f s e t dt

e dx x t

e dz z x ss

s

* Theorem 3.2 is a sufficient but not a necessary

condition.

＊ There may be different functions whose Laplace

transforms are the same

e.g., and

have the same Laplace transform

○ Theorem 3.3: Lerch’s Theorem

＊ Table 3.1 lists Laplace transforms of functions

9

( ) tf t e3

( )0 3

te tg t

t

, : continuous on 0,

If [ ] [ ], then

f g

L f L g f g

○ Theorem 3.1: Laplace transform is linear

Proof:

○ Definition 3.3:. Inverse Laplace transform

e.g.,

＊ Inverse Laplace transform is linear

10

[ ]( ) ( ) ( ),

, : real numbers

L f g s F s G s s a

0

0 0

[ ]( ) ( ( ) ( ))

( ) ( )

( ) ( )

st

st st

L f g s e f t g t dt

e f t dt e g t dt

F s G s

1 1( ) ( ( )) ( )

2

a i st

a ig t L G s G s e ds

i

1 12

1 1( ) , ( ) sin

1atL t e L t t

s a s

1 1 1[ ] [ ]L F G L F L G f g

3.2 Solution of Initial Value Problems Using Laplace Transform

○ Theorem 3.5: Laplace transform of

f: continuous on

: PC on [0, k]

Then, ------(3.1)

11

f

[0, )

f > 0k

lim ( ) 0 if 0sk

ke f k s

[ ]( ) ( ) (0)L f s sF s f

Proof:

Let

12

00 0( ) ( ) | ( )

k kst st k ste f t dt e f t se f t dt

0

0

[ ]( ) lim ( )

lim ( ) (0) ( )

k st

k

ksk st

k

L f s e f t dt

e f k f s e f t dt

(Integration by parts)udv uv vdu ',stu e dv f dt

0

Given

lim ( ) 0 ( ) (0)

0

sk st

ke f k s e f t dt f

s

( ) (0)sF s f

○ Theorem 3.6: Laplace transform of

: PC on [0, k]

for s > 0, j = 1,2 … , n-1

13

nf

1, ,..., : continuous on 0, nf f f

( )nf k

( )lim ( ) 0sk j

ke f k

( ) 1 2

( 2) ( 1)

[ ]( ) ( ) (0) (0)

(0) (0)

n n n n

n n

L f s s F s s f s f

sf f

。 Example 3.3:

From Table 3.1, entries (5) and (8)

14

4 1, (0) 1y y y

[ 4 ]( ) [ ]( ) 4 [ ]( ) ( ( ) (0)) 4 ( )

( 4) ( ) 1

L y y s L y s L y s sY s y Y s

s Y s

1[1]( )L s

s

1 1 1( 4) ( ) 1 , ( )

4 ( 4)s Y s Y s

s s s s

1 1 -1 11 1 1 1[ ] = L

4 ( 4) 4 ( 4)y L Y L L

s s s s s s

4 4 41 5 1( ) ( 1)

4 4 4t t ty t e e e

1 1(5) [ ] , (8) [ ]

( )( )

at btat e e

L e Ls a a b s a s b

15

Substitute into 4 1y y

4 4 45 1 5From , 4 5

4 4 4t t ty e y e e

4 4 4 45 5 14 4 4( ) 5 5 1 1

4 4 4t t t ty y e e e e

○ Laplace Transform of Integral

16

0

1[ ( ) ]( ) ( )

tL g x dx s G s

s

0

0

0

0

Let ( ) ( ) .

Then ( ) ( ) and (0) ( ) 0

( ) ( )( ) [ ( ) ]( )

t

t

f t g x dx

f t g t f g x dx

F s L f s L g x dx s

[ ] ( ) (0)L f L g G sF f sF

0

1( )

tF G L g x dx

s

From Eq. (3.1),

3.3. Shifting Theorems and Heaviside Function

3.3.1.The First Shifting Theorem

◎ Theorem 3.7:

○ Example 3.6: Given

17

[ ( )]( ) ( )atL e f t s F s a

0

( )

0 0

Proof: [ ( )]( ) ( )

( ) ( )

( ) ( )

at at st

s a t s t

L e f t s e e f s ds

e f t dt e f t dt

F s F s a

2 2[cos ]

sL bt

s b

2 2 [ cos ]

[( ) ]at s a

L e bts a b

○ Example 3.8:

18

1 12 2

4 4

4 20 ( 2) 16L L

s s s

2 2 2

4 [sin ] , [sin 4 ]

16

a xL at L t

s a s

22

4[ sin 4 ]

( 2) 16tL e t

s

1 22

4 sin 4

( 2) 16tL e t

s

3.3.2. Heaviside Function and Pulses ○ f has a jump discontinuity at a, if

exist

and are finite but unequal

○ Definition 3.4: Heaviside function

19

lim ( ) and lim ( )t a t a

f t f t

0 0( )

1 0

tH t

t

0a 0

( )1

t aH t a

t a

。

Shifting

20

0 ( ) ( )

( )

t aH t a g t

g t t a

。 Laplace transform of heaviside function

0

00

[ ( )]( ) ( )

1 1

st

st st

L H t s e H t dt

e dt es s

0[ ( )]( ) ( )

1

st

asst st

aa

L H t a s e H t a dt

ee dt e

s s

3.3.3 The Second Shifting Theorem ◎ Theorem 3.8:

Proof:

21

( ) ( ) ( ) ( )asL H t a f t a s e F s

0( ) ( )as sw ase e f w dw e F s

( ) ( ) ( )L H t a f t a s 0

( ) ( )ste H t a f t a dt

( )st

ae f t a dt

( )

0( )s a we f w dw

Let w t a t w a dt dw

○ Example 3.11:

Rewrite

22

2

0 0 2( )

1 2

tg t

t t

2( ) ( 2)( 1)g t H t t 2 2 21 ( 2 2) 1 ( 2) 4( 2) 5t t t t

2( ) ( 2)( 2) 4 ( 2)( 2) 5 ( 2)g t H t t H t t H t

2[ ] [ ( 2)( 2) ] 4 [ ( 2)( 2)] 5 [ ( 2)]L g L H t t L H t t L H t

2 2 2 2

23 2

[ ] 4 [ ] 5 [1]

2 4 5[ ]

s s s

s

e L t e L t e L

es s s

◎ The inverse version of the second shifting theorem

○ Example 3.13:

23

1[ ( )]( ) ( ) ( )asL e F s t H t a f t a

4 ( ), (0) (0) 0y y f t y y 0 3

( )3

tf t

t t

( ) ( 3)f t H t t

[ 4 ] [ ( )] [ ( 3) ]L y y L f t L H t t

3 3 3 32

[ ( 3) ] [ ( 3)( 3 3)] [ ( 3)( 3)] 3 [ ( 3)]

1 3 [ ] 3 [1] ----- (A)s s s s

L H t t L H t t L H t t L H t

e L t e L e es s

2

2 2

[ 4 ] [ ] 4 [ ] ( ) (0) (0) 4 ( )

( ) 4 ( ) ( 4) ( ) ----- (B)

L y y L y L y s Y s sy y Y s

s Y s Y s s Y s

2 3 3

2

1 3( ) ( ) ( 4) ( ) s sA B s Y s e e

ss

rewritten aswhere

24

2 2 2 2 2

3 1 3 1 3 1 1 1 1

( 4) 4 4 4 4 4 4

s s

s s s s s s

1 1 32 2

1 3 3 3 32 2 2

3 1( ) [ ( )]

( 4)

3 1 3 1 1 1 1

4 4 4 4 4 4

s

s s s s

sy t L Y s L e

s s

sL e e e e

s s s s

3 3

( 3) 1 ( 3)cos2( 3)4 4

1 1 1 + ( 3)( 3) ( 3) sin 2( 3)

4 4 2

H t H t t

H t t H t t

3 32 2 2 2 2

1 3 3 1( ) ( )

( 4) ( 4) ( 4)s ss

Y s e es s s s s s

25

0 3

3 3 1 1cos2( 3) ( 3) sin 2( 3) 3

4 4 4 8

t

t t t t

0 3

1[2 6cos2( 3) sin 2( 3)] 3

8

t

t t t t

3.4. Convolution

26

0( )( ) ( ) ( )

tf g t f t g d

◎ Theorem 3.9: Convolution theorem

Proof:

27

[ ] [ ] [ ]L f g L f L g F G

0 0( ) ( ) ( ) ( ) ( ) ( )st sF s G s F s e g t dt F s e g d

( ) [ ( ) ( )]se F s L H t f t

0( ) ( ) [ ( ) ( )] ( )F s G s L H t f t g d

0 0

0 0

0

( ) ( ) ( )

= ( ) ( ) ( )

= ( ) ( )

st

st

st

e H t f t dt g d

e g H t f t dtd

e g f t dtd

0 0 0 0

0

= ( ) ( ) = ( ) ( )

= ( * )( ) [ * ]( )

t tst st

st

e g f t d dt e g f t d dt

e f g t dt L f g s

◎ Theorem 3.10:

○ Exmaple 3.18

28

1[ ] *L FG f g

1 1 12 2

1 1 1[ ( ) ( )]

( 4) ( 4)L L L F s G s

s s s s

1 1 42

1 11 ( ), ( )

( 4)tL f t L te g t

s s

1 42

1 ( )* ( ) 1*

( 4)tL f t g t te

s s

4 4 4

0/ 4 /16 1/16

t t te d te e * *f g g f

0( * )( ) ( ) ( )

tf g t f t g d

0 t

0 ( ) ( )( 1) = ( ) ( ) ( * )( )

tf z g t z dz f z g t z dz g f t

◎ Theorem 3.11:

Proof :

Let z t

○ Example 3.19:

29

2 8 , (0) 1, '(0) 0y y y f y y

2[ 2 8 ] ( ( ) ) 2( ( ) 1)

8 ( ) [ ]( ) ( )

L y y y s Y s s sY s

Y s L f s F s

2

2 2

( 2 8) ( ) 2 ( )

1 2( ) ( )

2 8 2 81 1 1 1 1 1 2 1

( ) ( )6 4 6 2 3 4 3 2

s s Y s s F s

sY s F s

s s s s

F s F ss s s s

4 2 4 21 1 1 2( ) * ( ) * ( )

6 6 3 3t t t ty t e f t e f t e e

3.5 Impulses and Dirac Delta Function○ Definition 3.5: Pulse

○ Impulse:

○ Dirac delta function:

30

0

( ) ( ) 1 ,

0

t a

H t a H t b a t b a b

t b

1( ) [ ( ) ( )]t H t H t

0( ) lim ( )t t

1,t dt 0,t t 0

A pulse of infinite magnitude over an infinitely short duration

○ Laplace transform of the delta function

◎ Filtering (Sampling)

○ Theorem 3.12: f : integrable and continuous at a

31

1( ) [ ( ) ( )]t H t H t

1 1 1 1 1[ ( )] [ ( ) ( )]

ss e

L t L H t H t es s s

0 0 0 0

0

1[ ( )] [ lim ( )] lim [ ( )] lim lim

lim 1

s s

s

e seL t L t L t

s s

e

0( ) ( ) ( )f t t a dt f a

32

1[ ( ) ( )]

1[ ( ) ( ) ]

1 1[ ] [ ( ) ( ) ]

1 1[ ( ) ( ) ( ( ) ( ))] [ ( ) ( )]

a a

f t H t a H t a dt

f t H t a dt f t H t a dt

f t dt f a dt F t F ta a

F F a F F a F a F a

1[ ( ) ( )]f t H t a H t a da

Proof: ( )f t t a dt

33

0

1lim [ ( ) ( )]F a F a

0

lim ( )f t t a dt f t t a dt

0 0lim ( ) lim ( ) ( )F a f a f a

by Hospital’s rule

○ Example 3.20: 2 2 ( 3), (0) (0) 0y y y t y y

[ 2 2 ] [ ( 3)]L y y y L t 2 3( ) 2 ( ) 2 ( ) ss Y s sY s Y s e

33

2 2

1( )

2 2 ( 1) 1

sse

Y s es s s

12 2

1 1 sin , sin

1 ( 1) 1tL t L e t

s s

1 3 ( 3)2

1( ) ( 3) sin( 3)

( 1) 1s ty t L e H t e t

s

3.6 Laplace Transform Solution of Systems○ Example 3.22

Laplace transform

Solve for

34

2 3 2 4, (0) = (0) = (0) = 0

2 3 0

x x y yx x y

y x y

2 42 3 2

2 3 0

s X sX sY Ys

sY sX Y

( ) and ( )X s Y s

2

4 6 2( ) , ( )

( 2)( 1) ( 2)( 1)

sX s Y s

s s s s s s

( ) and ( )x t y t

Partial fractions decomposition

Inverse Laplace transform

35

2

7 1 1 1 1 10 1( ) 3

2 6 2 3 1X s

s s s s

1 1 1 2 1

( )3 3 2 3 1

Y ss s

1 27 1 10( ) ( ) 3

2 6 3t tx t L X t e e

1 21 2( ) ( ) 1

3 3t ty t L Y e e

3.7. Differential Equations with Polynomial Coefficient

◎ Theorem 3.13:

Proof:

○ Corollary 3.1:

36

[ ( )]( ) ( )L tf t s F s

0 0( ) ( ) ( )

ststd de

F s e f t dt f t dtds ds

0( ) stte f t dt

[ ( )]( )L tf t s( )[ ( )]( ) ( 1) ( )n n nL t f t s F s

○ Example 3.25:

Laplace transform

37

(4 2) 4 0, (0) 1ty t y y y

[ ] 4 [ ] 2 [ ] 4 [ ] 0 ---- ( )L ty L ty L y L y A 2

2

[ ] [ ] [ (0) (0)]

= 2 1

d dL ty L y s Y sy y

ds ds

sY s Y

[ ] [ ] [ (0)]d d

L ty L y sY y Y sYds ds

[ ] (0) 1L y sY y sY 2( ) 2 1 4 4 2 2 4 0A sY s Y Y sY sY Y

4 8 3 ------ ( )

( 4) ( 4)

sY Y B

s s s s

Find the integrating factor,

Multiply (B) by the integrating factor

38

2 24 8

ln[ ( 4) ] 2 2( 4) ( 4)s

dss ss se e s s

2 2( 4) (4 8) ( 4) 3 ( 4)s s Y s s s Y s s

2 2[ ( 4) ] 3 ( 4)s s Y s s 3 2

2 2 3 22 2

6( 4) 6 , ( )

( 4)

s s cs s Y s s c Y s

s s

Inverse Laplace transform

39

2 2 2

2 2 2

2 2

2 2

6( ) =

( 4) ( 4) ( 4)

4 4 6

( 4) ( 4) ( 4)

1 2

4 ( 4) ( 4)

1 2 1 4 1( )

4 ( 4) 16 4

s cY s

s s s s

s c

s s s s

c

s s s s

c

s s s s s

4 4 4 4( ) 2 [ 1 2 2 ]32

t t t tcy t e te t e te

○ Apply Laplace transform to

algebraic expression for Y

Apply Laplace transform to

Differential equation for Y

40

( ), , : constanty Ay By f t A B

( ) ( ) ( )y p t y g t y f t

◎ Theorem 3.14: PC on [0, k],

41

0k , , s.t. ( ) , 0btM b f t Me t

lim ( ) 0s

F s

0 0

( )

0 0

Proof :

0 ( ) ( ) ( )

-

0 lim ( ) lim 0

lim ( ) 0 and hence lim ( ) 0

st st

st bt s b t

s s

s s

F s e f t dt e f t dt

M Me Me dt e

b s s b

MF s

s bF s F s

○ Example 3.26:

Laplace transform

------(A)

------(B)

42

2 4 1, (0) (0) 0y ty y y y

2 ( ) (0) (0)

1 2 [ ]( ) 4 ( )

s Y s sy y

L ty s Y ss

[ ]( ) ( [ ]( ))

( ( ) (0)) ( ) ( )

dL ty s L y s

dsd

sY s y Y s sY sds

2( ) ( ) 2 ( ) 2 ( ) 4 ( ) 1/A s Y s Y s sY s Y s s

2

3 1( )

2 2

sY Y

s s

Finding an integrating factor,

Multiply (B) by ,

43

2 2

3ln2 34 43 1

( ) 3ln ,2 4

s sss

ds s s e s es

2

3 4

s

s e

2 2 2

3 3 34 4 42

3 1( )

2 2

s s sss e Y s e Y s e

s s

2 2 2

24

3 4 4 4(3 )2 2

s s ss ss e Y s e Y e

In order to have

44

2

43 3

1( )

scY s e

s s

2 2

3 4 4( )2

s sss e Y e

2 2 2

3 4 4 41

2

s s s

s e Y se e c

lim ( ) 0,s

Y s

23

1 1Choose 0, ( ) and ( )

2c Y s y t t

s

Formulas:

○ Laplace Transform:

○ Laplace Transform of Derivatives:

○ Laplace Transform of Integral:

45

0[ ]( ) ( )stL f s e f t dt

[ ]( ) ( ) (0)L f s sF s f

( ) 1 2

( 2) ( 1)

[ ]( ) ( ) (0) (0)

(0) (0)

n n n n

n n

L f s s F s s f s f

sf f

0

1[ ( ) ]( ) ( )

tL g x dx s G s

s

○Shifting Theorems:

○ Convolution:

Convolution Theorem:

○

46

[ ( )]( ) ( )atL e f t s F s a

( ) ( ) ( ) ( )asL H t a f t a s e F s

0( )( ) ( ) ( )

tf g t f t g d

[ ]L f g F G

( )[ ( )]( ) ( ), [ ( )]( ) ( 1) ( )n n nL tf t s F s L t f t s F s

○

47

1[0] 0, [1]L L

s

2 1

2 3

1 2[ ] , [ ] , [ ] !/n nL t L t L t n s

s s

2 2 2

1[sin ] , [sin ]

1

aL t L at

s s a

2 2 2[cos ] , [cos ]1

s sL t L at

s s a

1 1[ ] ,

( )( )

at btat e e

L e Ls a a b s a s b

1[ ( )] , [ ( )]

aseL H t L H t a

s s

[ ( )] 1L t